Assertion: If $\frac{\text{dy}}{\text{dy}}+\text{xy}=\text{x}^3\text{y}^3,\text{x}>0,\text{y}\geq0$ and $\text{y}(0)=1,$ then $\text{y}(1)=\frac{1}{\sqrt{2}}$
Reason: The differential equation is linear with integrating factor ex
- Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- Assertion is correct statement but Reason is wrong statement.
- Assertion is wrong statement but Reason is correct statement.
- Assertion is correct statement but Reason is wrong statement.
Solution:
$\frac{1}{\text{y}^3}\frac{\text{dy}}{\text{dx}}+\frac{\text{x}}{\text{y}^2}=\text{x}^3$
Put $\frac{1}{\text{y}^2}=\text{z}\Rightarrow\frac{2}{\text{y}^3}\text{dy}=\text{dz}$
$\therefore\frac{\text{dz}}{\text{dx}}-2\text{xz}=-2\text{x}^3,$
which is a linear differential equation with $\text{I.F}=\text{e}^{\text{x}^2}$
$\therefore$ $\text{ze}^{-\text{x}^2}=-\int\text{e}^{\text{x}^2}2\text{x}^3\text{dx} $
$\Rightarrow\text{ze}^{-\text{x}^2}=(\text{x}^2+1)\text{e}^{\text{-x}^2}+\text{C}\Rightarrow\text{z}=\text{x}^2+1+\text{C}\text{e}^{\text{x}^2}$
$\therefore\frac{1}{\text{y}^2}=\text{x}^2+1+\text{C}\text{e}^{\text{x}^2}$
$\because\text{y}(0)=1\Rightarrow\text{c}=0$
$\therefore\text{y}^2=\frac{1}{\text{x}^2+1}\Rightarrow\text{y}=\frac{1}{\sqrt{\text{x}^2+1}}\Rightarrow\text{y}(1)=\frac{1}{\sqrt{2}}$